System and method for cable resistance cancellation

ABSTRACT

A method for cable resistance cancellation. A single remote sense line and a simple cable resistance cancellation network are leveraged in a power supply unit to compensate for the total cable voltage drop, while maintaining tight output accuracy. By completely compensating for the voltage drops, the wire gauge for the main power wires can be reduced, thereby allowing the use of smaller diameter cables.

FIELD OF THE INVENTION

The present invention is generally related to power supply cables, andmore specifically to power supply cable resistance cancellation.

BACKGROUND OF THE INVENTION

For an external power supply, the output cable wire resistance willcause load regulation issues, as the current drawn from the load varies.Cable wire gauge and length will also affect the output voltage accuracyas the resistance of the load changes. Traditionally, in order tomaintain tight output accuracy, the output cable and circuits haveincorporated two remote sense wires which are able to compensate for thetotal cable voltage drop. Another technique used is lowering the gage(increasing the diameter) of the output wire to lessen the effects ofthe voltage drop. Lack of voltage drop compensation means that outputcables have to be sized with large AWG wire to minimize the IR drops andthat open circuit (no load) output voltage had to be programmed high tominimize the effects of the drops. Also, output cables had to be limitedto a maximum length for a given wire gauge so the drops at high currentdid not become excessive.

Solutions using an interchangeable tip connector can accommodate forcable voltage drops, but require two remote sense wires in the cable,leading to two additional connector pins in the tip, to achieve thevoltage correction. Adding a second remote sense wire and two more tipconnector pins increase the weight and size of the cable and tipconnector, thereby adding unneeded bulk and weight for a user. A singlewire solution would have only compensated for half the drop passivelyand would have required an active stage to compensate for the entiredrop on the positive and negative leads. The active compensation wouldhave placed a length restriction on an output cable as well.

Voltage programming, as done in Mobility Electronics'Juice 70 productline, uses a resistor in a tip which is connected in parallel with avoltage sampling potentiometer chain in a base unit. The tip resistorshares a common ground with the high current ground return, so any loadtransients in the cable ground lead are transmitted back to the baseunit and affect the output voltage. The voltage programming was limitedto a pre-set minimum voltage defined by the reference in the base unit.This meant that no passive tip could force the base unit output voltagebelow the preset minimum (thus, no ability to upgrade the power supplyto operate over a wider voltage range other than changing the baseunit). The preset minimum on the Juice products is 15 volts. The voltageprogramming resistor in the tip was included in the overall feedbackloop along with the cabling from the base unit to the programmingresistor. This puts a portion of the feedback loop outside the baseunit, making loop compensation more difficult.

FIG. 1 discloses a 4-wire prior art example of a voltage programmingsystem 100. In this example, note that load current, I_(L), flowsthrough the ground path shared by R₁₂₂. Noise or transients on thisground or picked up by the Vsense line cannot be filtered by theaddition of bypass capacitors (across R₁₁₄ or R₁₂₂) as this would placean extra pole on integrator U1 outside its local feedback loop. Thisexternal low frequency pole would make loop compensation difficult anddegrade the transient response of the base unit. Also, the distributedcapacitance of the cable itself produces a similar pole at higherfrequencies, again contributing to problems in compensating the feedback loop.

SUMMARY OF INVENTION

The present invention achieves technical advantages as a method forcable resistance cancellation. One embodiment of the invention utilizesa single remote sense line and a simple cable resistance cancellationnetwork in a power supply unit to compensate for the total cable voltagedrop, while maintaining tight output accuracy. By completelycompensating for the voltage drops, the wire gauge for the main powerwires can be reduced, thereby allowing the use of smaller diametercables.

BRIEF DESCRIPTION OF THE DRAWINGS

FIG. 1 is a diagram of a prior art voltage programming system;

FIG. 2 is a diagram of an output voltage programming system inaccordance with an exemplary embodiment of the present invention; and

FIG. 3 is a diagram of an output current limiting system in accordancewith an exemplary embodiment of the present invention.

DETAILED DESCRIPTION OF THE PRESENT INVENTION

Referring now to FIG. 2, there is shown at 200 a diagram of an outputvoltage programming system in accordance with an exemplary embodiment ofthe present invention. Voltage programming system 200 compensates forcable voltage drop, while maintaining tight voltage output accuracy.Voltage programming system 200 is comprised of base unit 202, cable 204,and tip 206.

Base unit 202 contains a standard voltage comparator/integratoramplifier U1 with feedback from the internal Vout signal. The Voutfeedback is attenuated by the resistor network of R1 and R2. The otherinput to this amplifier is from a resistor divider in tip 206. The tipresistor divider sets the output voltage for the tip output connector.Also in base unit 202 is voltage reference, Vref, that is used toprovide a current to tip 206 in order to establish 2.5V at the top ofthe tip resistor divider. This current will vary as discussed laterbased on the total value of the tip resistor divider. Base unit 202 hasa base resistor divider composed of R and kR. These two resistorsprovide the compensation for the cable voltage drops as output loadcurrent varies. They are typically low resistance value, 1% toleranceresistors. Note that the R and kR resistors are connected across analogground 220 and the internal power ground 222. This connection providesthe sampling of cable voltage drop that allows the compensation to work.

Cable 204 connects base unit 202 to tip 206, and is made with 5 wires onthe adapter side and 4 wires on the tip 206 side. Pins 1 and 5 are heavyduty current carrying pins and the rest are low current signal pins. AKelvin connection is made in cable 204 at the tip 206 end from pin 4 topin 5. With the cable voltage drop compensation the voltage drop incable 204 in both directions is completely eliminated. This would seemto imply that the usual heavy gauge wire that is used in the power leadscould be reduced significantly to make cable 204 smaller. Care must beused in sizing the power leads so that the loss in cable 204 does notdegrade the overall adapter and cable efficiency so much that the CECand EnergyStar requirements cannot be met. Also, any cable 204 voltagedrop compensation is made up in tip 206 itself with a higher Voutvoltage. This means that the output voltage range of base unit 202 musttake this higher compensating output voltage into account, in circuitand magnetics design.

Tip 206 contains the two resistors used to set the output voltage andoutput current limiting. The current limiting is set by computing thetotal value (Ri) of the two resistors in the resistor divider made up ofnRi and (1−n)Ri. The output voltage of the Tip is set by controlling theratio of the two resistors by the factor n, where n<1.

In this new architecture the voltage and current limit programming sharea common resistance in the tip labeled Ri. The total value of Ri definesthe current limit point and the ratio of the upper and lower resistorsthat make up Ri define the adapter output voltage. For this explanationof voltage programming, the voltage at the top of Ri will be assumed tobe equal to an internal adapter reference called Vref equal to 2.5volts. As will be shown, the value of Vref can be changed as long as thevalues of R1, R2, R and kR are changed as well.

To program the Tip output voltage (compensated voltage delivered to theTip output pins) the following equation is used:

$\begin{matrix}{{Vout}^{\prime} = {\frac{\left( {{R\; 1} + {R\; 2}} \right)}{R\; 2} \cdot ({Vref}) \cdot n}} & {{Equation}\mspace{20mu} A}\end{matrix}$

Where R₁ and R₂ are fixed by the architecture and the variable n definesthe output voltage. Referring to FIG. 2. resistor R₁ is set to 102K Ωand R₂ to 11.3K Ω.

The R₁ and R₂ values chosen are the nearest 1% E96 values that produce avoltage division ratio closest to 10. In this case the 102K Ω and 11.3KΩ give a ratio of 10.026549. For the purpose of this derivation thatnumber will be rounded down to 10. The choice of 10 for a ratio is madeso that in case of a loss of sense ground in the tip the maximum outputvoltage for n=1 would be limited to 25 v. This would not exceed thetypical over voltage protection level of 28 volts on most powersupplies. Substituting in 10 for the ratio n in Equation A yields thefollowing equation:Vout′=10·(2.5 v)·n  Equation B

In order to produce an output voltage in the range of 12 volts to 24volts, the variable n must have a value in the range of 0.48 to 0.96,respectively. This n value is independent of the total value of R₁ inthe tip. The compensation resistors R and kR are controlled by thevalues of R₁ and R₂. The resulting values for R and kR are 14.7Ω and118Ω respectively. These values are chosen to be low so they do notappreciably affect the voltage at the junction of R₁ and R₂.

Referring now to FIG. 3, there is shown at 300 a diagram of an outputcurrent limiting system in accordance with an exemplary embodiment ofthe present invention.

The limiting current is set by the total value of Ri, where Ri isdefined by:

$\begin{matrix}{{Ri} = \frac{{Vref} \cdot G^{- 1}}{Ilimit}} & {{Equation}\mspace{20mu} C}\end{matrix}$

wherein Vref=2.5 V, G=100 uA/A, and Ri=R_(a)+R_(b).

The G term defines the transfer function of the current mirror composedof Q1-Q4. This transfer function is fixed within the architecture, butthe Vref, conversion gain of the current mirror defined by R₈ and thegain F[G] of the current sensor on the base unit can all be variables aslong as the G term remains at 100 uA/A.

This demonstrates the development of the formulas for the voltage dropcompensation provided by R and kR. Also, the dependency of R and kR onR1 and R2 is shown. On the inverting input of U1 a voltage Vp isdeveloped from Vout, and the network of R1, R2, R, kR and the returnload current It, flowing in one leg of the output cable between pins 5′and 5. The purpose of the compensation network is to sample the current(I_(load)) returning from the load, convert that current into a voltage(I_(load)*Rcable) and apply a portion of that voltage to raise thevoltage Vp proportional to the load current. By raising Vp, the Voutvoltage will be raised to offset the voltage drops in the outputcabling. The equations for Vout and Vp are derived below and thesubsequent compensation of the tip voltage Vout for cable drops due toload current Iload is shown.

Care is needed in picking the R value used in the compensation network.Since k is a dimensionless and constant, any starting value of R couldbe used. However, in order to keep errors to a minimum in the system,some guidelines for R are presented.

First, because (R+kR) is effectively in parallel with the cableresistance, rcable, the value of R (and subsequently kR) should not beso low as to carry appreciable branch current. A good rule of thumb isfor R+kR to be three orders of magnitude greater than the largest singleleg resistance expected in the cable and connectors. For example, for a6 foot output cable with 18 AWG wire and two connectors the totalresistance could be: (0.03 ohms/connector*2 connectors)+(6 feet*0.006ohms/foot)=0.096 ohms. Rounding up to 0.1 ohms yields a minimum value of1000*0.1 ohms=100 ohms for R+kR. At a 10 amp load current the branchcurrent in R+kR is only 10 mA, thus 0603 resistors can be used and themeasurement error due to the branch current stealing is only 0.1%.

At low currents the compensation effects of R and kR are minimal.However, the parallel combination kR and (R+Rcable) is effectively inseries with the lower voltage setting resistor R2. If It is chosen toolarge this added resistance will increase the apparent resistance of R2and this will consequently cause the ratio of R1 and R2 to be in error.This error will result in a lower than expected output voltage. A goodrule of thumb is to keep the parallel combination of R and kR less than1% of the value of R2 (Rcable can be ignored since it is very small).For example, if R2=11.3K Ω, then R∥kR should be less than 113Ω. For thevalues of R=14.7Ω and kR=118Ω the parallel combination is <14.7Ω whichsatisfies the requirement. In this case the proportion of the R and kRcombination to R2 is 13.1 Ω/11.3K Ω=0.12% of R2.

$\begin{matrix}{{{Vp} = {{m \cdot \left( {{Vout} - {Vr}} \right)} + {Vr}}}{{{where}\mspace{14mu} m} = \frac{R\; 2}{{R\; 1} + {R\; 2}}}} & {{Equation}\mspace{20mu} 1}\end{matrix}$

When the system is in balance the negative and positive terminals of U1are equal and this leads to:Vp=m·(Vout−Vr)+Vr=V _(VP) +Iload·Rcable  Equation 2

Vr is relative to the internal GND point and Vvp is relative to the pin5′ ground in the tip.

Substituting for m in Equation 1, yields:

$\begin{matrix}{{{{Vp} = {{\frac{R\; 2}{{R\; 1} + {R\; 2}} \cdot {Vout}} + {{Vr} \cdot \left( {1 - \frac{R\; 1}{{R\; 1} + {R\; 2}}} \right)}}},{but}}{{{1 - \frac{R\; 2}{{R\; 1} + {R\; 2}}} = \frac{R\; 1}{{R\; 1} + {R\; 2}}},{thus}}} & {{Equation}\mspace{20mu} 3} \\{{{Vp} = {{\frac{R\; 2}{{R\; 1} + {R\; 2}} \cdot {Vout}} + {{Vr} \cdot \frac{R\; 1}{{R\; 1} + {R\; 2}}}}},{{{and}\mspace{14mu}{Vr}} = {\frac{kR}{R + {kR}} \cdot {Iload} \cdot {Rcable}}}} & {{Equation}\mspace{20mu} 4}\end{matrix}$

Substituting for Vr in Equation 4, yields:

$\begin{matrix}{{Vp} = {{\frac{R\; 2}{{R\; 2} + {R\; 2}} \cdot {Vout}} + {\frac{R\; 1}{{R\; 1} + {R\; 2}} \cdot \left( {\frac{kR}{R + {kR}} \cdot {Iload} \cdot {Rcable}} \right)}}} & {{Equation}\mspace{20mu} 5}\end{matrix}$

and from Equation 2, Equation 5 becomes:

$\begin{matrix}\begin{matrix}{{Vp} = {{\frac{R\; 2}{{R\; 1} + {R\; 2}} \cdot V_{out}} + {\frac{R\; 1}{{R\; 1} + {R\; 2}} \cdot}}} \\{\left( {\frac{kR}{R + {kR}} \cdot {Iload} \cdot {Rcable}} \right)} \\{= {V_{VP} + {{Iload} \cdot {Rcable}}}}\end{matrix} & {{Equation}\mspace{20mu} 6}\end{matrix}$

Rewriting Equation 6 gives:

$\begin{matrix}{{Vp} = {{\frac{R\; 2}{{R\; 1} + {R\; 2}} \cdot {Vout}} + {\frac{R\; 1}{{R\; 1} + {R\; 2}} \cdot \left( {\frac{kR}{R\left( {1 + k} \right)} \cdot {Iload} \cdot {Rcable}} \right)}}} \\{= {{Vvp} + {{Iload} \cdot {Rcable}}}}\end{matrix}$

Through simplification:

$\begin{matrix}{{Vp} = {{\frac{R\; 2}{{R\; 1} + {R\; 2}} \cdot {Vout}} + {\frac{R\; 1}{{R\; 1} + {R\; 2}} \cdot \left( {\frac{k}{1 + k} \cdot {Iload} \cdot {Rcable}} \right)}}} \\{= {{Vvp} + {{Iload} \cdot {Rcable}}}}\end{matrix}$

Dropping V_(p) and setting the solution in terms of Vout yields:

$\begin{matrix}{{{Vout} \cdot \frac{R\; 2}{{R\; 1} + {R\; 2}}} = {{Vvp} + {{Iload} \cdot {Rcable}} - \left( {{\frac{R\; 1}{{R\; 1} + {R\; 2}} \cdot \frac{k}{k + 1}}{{Iload} \cdot {Rcable}}} \right)}} & {{Equation}\mspace{20mu} 7}\end{matrix}$

Now, rearranging Equation 7 to find Vout:

$\begin{matrix}{{Vout} = {{{Vvp}\frac{{R\; 1} + {R\; 2}}{R\; 2}} + {{Iload} \cdot {Rcable} \cdot \frac{{R\; 1} + {R\; 2}}{R\; 2}} - {\left( {\frac{R\; 1}{{R\; 1} + {R\; 2}} \cdot \frac{k}{k + 1} \cdot {Iload} \cdot {Rcable}} \right) \cdot \frac{{R\; 1} + {R\; 2}}{R\; 2}}}} & {{Equation}\mspace{20mu} 8}\end{matrix}$

or, finally:

$\begin{matrix}{{Vout} = {{{Vvp} \cdot \frac{{R\; 1} + {R\; 2}}{R\; 2}} + {{Iload} \cdot {Rcable} \cdot \left( {\frac{{R\; 1} + {R\; 2}}{R\; 2} - {\frac{R\; 1}{R\; 2} \cdot \frac{k}{k + 1}}} \right)}}} & {{Equation}\mspace{20mu} 9}\end{matrix}$

Equation 9 now describes the V_(out) voltage inside the adapter brick.This voltage, while interesting, is not of primary concern. The voltageat the end of the cable is most important to regulate and the followingderivations show how the compensation is achieved.

The Vout′ voltage at the cable end is described by:Vout′=Vload==Vout−(2·Rcable)·Iload  Equation 10

Substituting Equation 9 for Vout leads to:

$\begin{matrix}\begin{matrix}{{Vout}^{\prime} = {Vload}} \\{= {{V_{VP} \cdot \frac{{R\; 1} + {R\; 2}}{R\; 2}} + {{Iload} \cdot}}} \\{{{{Rcable}\left( {\frac{{R\; 1} + {R\; 2}}{R\; 2} - {\frac{R\; 1}{R\; 2} \cdot \frac{k}{k + 1}}} \right)} -}\;} \\{\left( {2 \cdot {Rcable}} \right) \cdot {Iload}}\end{matrix} & {{Equation}\mspace{20mu} 11}\end{matrix}$

And simplifying again gives:

$\begin{matrix}{{Vout}^{\prime} = {Vload}} \\{= {{V_{vp} \cdot \frac{{R\; 1} + {R\; 2}}{R\; 2}} + {{Iload} \cdot {Rcable} \cdot}}} \\{\left\lbrack {\frac{{R\; 1} + {R\; 2}}{R\; 2} - 2 - \left( {\frac{R\; 1}{R\; 2} \cdot \frac{k}{k + 1}} \right)} \right\rbrack}\end{matrix}$

A key observation is if the term inside the brackets can be made toequal zero, then the effects of Iload and Rcable are eliminated andVout′becomes independent of cable length and load current. Vvp isindependent of load current and only depends on the ratio of Ra and Rbin the tip. For this reason Vvp can be treated as a constant. Thus,setting the bracketed term=0 and re-arranging the parentheses yields:

$\begin{matrix}{\left\lbrack {\frac{{R\; 1} + {R\; 2}}{R\; 2} - 2 - \left( {\frac{R\; 1}{R\; 2} \cdot \frac{k}{k + 1}} \right)} \right\rbrack = 0} & {{Equation}\mspace{20mu} 12}\end{matrix}$

Re-arranging:

${\frac{{R\; 1} + {R\; 2}}{R\; 2} - 2} = {\frac{R\; 1}{R\; 2} \cdot \frac{k}{k + 1}}$

Euler's method is applied to the left side of the equation:

${\frac{{R\; 1} + {R\; 2}}{R\; 2} \cdot \frac{R\; 2}{R\; 2}} - {2\frac{R\; 2}{R\; 2}}$

This yields:

$\frac{\left( {{R\; 1} + {R\; 2}} \right) - {{2 \cdot R}\; 2}}{R\; 2}$

Which is:

$\frac{{R\; 1} - {R\; 2}}{R\; 2}$

Substituting back into Equation 12 yields:

$\begin{matrix}{\frac{{R\; 1} - {R\; 2}}{R\; 2} = {\frac{R\; 1}{R\; 2} \cdot \frac{k}{k + 1}}} & {{Equation}\mspace{20mu} 13}\end{matrix}$

Now multiplying both sides by R2 and moving R1 yields:

$\begin{matrix}{\frac{{R\; 1} - {R\; 2}}{R\; 1} = \frac{k}{k + 1}} & {{Equation}\mspace{20mu} 14}\end{matrix}$

Solving for k:

$\frac{R\; 1}{{R\; 1} - {R\; 2}} = {\frac{k + 1}{k} = {{\frac{k}{k} + \frac{1}{k}} = {\frac{1}{k} + 1}}}$so: ${\frac{R\; 1}{{R\; 1} - {R\; 2}} - 1} = \frac{1}{k}$

now with a lowest common denominator (LCD) on the left:

${\frac{R\; 1}{{R\; 1} - {R\; 2}} - \frac{{R\; 1} - {R\; 2}}{{R\; 1} - {R\; 2}}} = {\frac{{R\; 1} - {R\; 1} + {R\; 2}}{{R\; 1} - {R\; 2}} = \frac{1}{k}}$

yielding:

$\begin{matrix}{k = \frac{{R\; 1} - {R\; 2}}{R\; 2}} & {{Equation}\mspace{20mu} 15}\end{matrix}$

This k value represents the ratio of the two compensation resistors Rand kR. Once R1 and R2 are set, then k can be determined from Equation15. This cancellation network can be used with Mobility's programmingpower supply and any external power supply with 2 voltage sense wires.

As a check, this equation will be substituted back into Equation 11 fork, to verify the terms are equal to 0. From Equation 11:

${Vout}^{\prime} = {{Vload} = {{{Vvp} \cdot \frac{{R\; 1} + {R\; 2}}{R\; 2}} + {{Iload} \cdot {Rcable} \cdot \left\lbrack {\frac{{R\; 1} + {R\; 2}}{R\; 2} - 2 - \left( {\frac{R\; 1}{R\; 2} \cdot \frac{k}{k + 1}} \right)} \right\rbrack}}}$and $\frac{k}{k + 1} = \frac{{R\; 1} - {R\; 2}}{R\; 1}$

from Equation 14.

Substituting into Equation 11:

${Vout}^{\prime} = {{Vload} = {{{Vvp} \cdot \frac{{R\; 1} + {R\; 2}}{R\; 2}} + {{Iload} \cdot {Rcable} \cdot \left\lbrack {\frac{{R\; 1} + {R\; 2}}{R\; 2} - 2 - \left( {\frac{R\; 1}{R\; 2} \cdot \frac{{R\; 1} - {R\; 2}}{R\; 1}} \right)} \right\rbrack}}}$${so},{{Vout}^{\prime} = {{Vload} = {{{Vvp} \cdot \frac{{R\; 1} + {R\; 2}}{R\; 2}} + {{Iload} \cdot {Rcable} \cdot \frac{{R\; 1} + {R\; 2} - {{2 \cdot R}\; 2} - {R\; 1} + {R\; 2}}{R\; 2}}}}}$${But},{\frac{{R\; 1} + {R\; 2} - {{2 \cdot R}\; 2} - {R\; 1} + {R\; 2}}{R\; 2} = 0},{{{so}:{Vout}^{\prime}} = {{Vload} = {{Vvp}\frac{{R\; 1} + {R\; 2}}{R\; 2}}}}$

thereby verifying the original assumption.

Advantageously, the tip programming is achieved by first by selectingthe value of R1 that gives the desired Ilimit from Equation C, then thevalue of n from Equation A that defines the output voltage at the tip isselected. This n value defines the ratio of Ra to Rb in the tip.

The present invention derives technical advantages because first, othersolutions can't reduce the number of wires in the output cable to 5.Additionally, two of the wires are the same small AWG for currentcarrying and three wires are smaller with only signal voltages andcurrents. The base unit interface has 5 pins and the tip interfaceconnector has 4 pins. The voltage drops in the cable are 100%compensated for by the addition of a single 5th wire in the cable and 2resistors in the base unit. The output cable can be thinner since thevoltage drop in the cable is compensated in the base unit (thisadvantage must be balanced with the CEC and EnergyStar efficiencyrequirements).

The present invention achieves further technical advantages becausevoltage drop compensation is independent of cable length and wire gaugeand can reduce the error due to voltage drops by a factor of 100 ormore. Moreover, the tip components that program the output voltage andcurrent are now completely outside the feedback loop so loopcompensation is simplified and cable noise can be bypassed withoutcreating extra feedback poles. The output voltage range of a properlydesigned base unit could be programmed from its maximum rated output allthe way down to 0.0 volts with passive components in the tip. Highcurrents in the power or ground leads do not affect the loop transientresponse and have minimal affect on the output voltage programming.

Though the invention has been described with respect to a specificpreferred embodiment, many variations and modifications will becomeapparent to those skilled in the art upon reading the presentapplication. It is therefore the intention that the appended claims beinterpreted as broadly as possible in view of the prior art to includeall such variations and modifications.

1. A voltage programming circuit, comprising: a base unit adapted toprovide an output voltage; a cable adapted to couple the output voltagefrom the base unit to a remote load and create a load voltage; areference line, disposed within the cable, adapted to couple the loadvoltage at the load back to the base unit; a cable cancellation networkadapted to compensate for total voltage drop between the base unit andthe load due to the cable; and a tip adapted to operably and securelycouple the cable to the load; wherein the load voltage is:${{Vout}^{\prime} = {\frac{\left( {{R\; 1} + {R\; 2}} \right)}{R\; 2} \cdot ({Vref}) \cdot n}},$wherein Vout′ is the value of the load voltage; Vref is the value of areference voltage; R1 is the value of a first resistor; R2 is the valueof a second resistor; and n is the value of a voltage adjustment.
 2. Thevoltage programming circuit of claim 1, wherein the base unit is a powersupply.
 3. The voltage programming circuit of claim 1, wherein the loadis a portable electronic device.
 4. The voltage programming circuit ofclaim 1, wherein the cable cancellation network is a resistive network.5. The voltage programming circuit of claim 1, wherein the load voltageis a function of the cable cancellation network.
 6. The voltageprogramming circuit of claim 1, wherein the load voltage is set bycontrolling the value of n.
 7. The voltage programming circuit of claim6, wherein n<1.
 8. The voltage programming circuit of claim 1, whereinVref is 2.5V.
 9. The voltage programming circuit of claim 1, wherein thecancellation network, is defined by the equation:${\left\lbrack {\frac{R_{1} + R_{2}}{R_{2}} - 2 - \left( {\frac{R_{1}}{R_{2}} \cdot \frac{k}{k + 1}} \right)} \right\rbrack = 0},$wherein k is a compensation factor.
 10. The voltage programming circuitof claim 9, wherein $k = {\frac{R_{1} - R_{2}}{R_{2}}.}$
 11. A currentprogramming circuit, comprising: a base unit adapted to provide anoutput voltage to a load; a cable adapted to propagate the outputvoltage from the base unit to the load; a reference line, disposedwithin the cable, adapted to propagate the output voltage at the load tothe base unit; a cable cancellation network adapted to compensate fortotal voltage drop due to the cable; and a tip adapted to operably andsecurely engage the load and one end of the cable; wherein the cablecancellation network is defined by the equation:${\left\lbrack {\frac{R_{1} + R_{2}}{R_{2}} - 2 - \left( {\frac{R_{1}}{R_{2}} \cdot \frac{k}{k + 1}} \right)} \right\rbrack = 0},$wherein R1 is a first resistor value; R2 is a second resistor value; andk is a compensation factor.
 12. The current programming circuit of claim11, wherein the base unit is a power supply.
 13. The current programmingcircuit of claim 11, wherein the load is an electronic device.
 14. Thecurrent programming circuit of claim 11, wherein the cable cancellationnetwork is a resistive network.
 15. A voltage programming circuit,comprising: a base unit adapted to provide an output voltage; a cableadapted to couple the output voltage from the base unit to a remote loadand create a load voltage; a reference line, disposed within the cable,adapted to couple the load voltage at the load back to the base unit; acable cancellation network adapted to compensate for total voltage dropbetween the base unit and the load due to the cable; and a tip adaptedto operably and securely couple the cable to the load; wherein an outputcurrent is limited with a resistive value, which is a function of thecable cancellation network; and wherein the resistive value is:${{Ri} = \frac{{Vref} \cdot G^{- 1}}{I\mspace{20mu}{limit}}},$ wherein Ilimit is the value of the output current limit; Vref is the value of thereference voltage; G is the transfer function of a current mirror; andRi is the value of Ra+Rb, wherein Ra is the value of a first resistor inthe tip, and wherein Rb is the value of a second resistor in the tip.16. The voltage programming circuit of claim 15, wherein G is 100 uA/A.17. The voltage programming circuit of claim 15, wherein Vref is 2.5V.18. The voltage programming circuit of claim 15, wherein$k = {\frac{R_{1} - R_{2}}{R_{2}}.}$